= Solution
The constant $R$ is the curvature radius of the closed spatial slices when $a=1$; at general scale factor their curvature radius is $Ra$ and their scalar curvature is $6/(R^2a^2)$.
The <cosmological perfect-fluid continuity equation> is
$$
\dot\rho+3\frac{\dot a}{a}(\rho+P)=0.
$$
During expansion, $\dot a>0$, and therefore
$$
\frac d{dt}(\rho a^3)
=-3Pa^2\dot a\leq0.
$$
If $C=\rho(t_0)a(t_0)^3$, then throughout the later expanding phase
$$
\rho a^3\leq C.
$$
Multiplying the <Friedmann equation> by $a^2$ gives
$$
\dot a^2
=\frac{8\pi G}{3c^2}\rho a^2
-\frac{c^2}{R^2}
\leq
\frac{8\pi GC}{3c^2a}
-\frac{c^2}{R^2}.
$$
The right-hand side is negative once
$$
a>\frac{8\pi GCR^2}{3c^4},
$$
which is impossible because $\dot a^2\geq0$. Thus the scale factor cannot grow without bound.
In fact the expanding phase ends in finite time. Combining the Friedmann and continuity equations yields the <Friedmann acceleration equation>
$$
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)<0.
$$
The curvature term in the first Friedmann equation also implies
$$
\rho\geq\frac{3c^4}{8\pi GR^2a^2}.
$$
While $a$ remains below the finite bound just found, $\ddot a$ is bounded above by a strictly negative constant. Hence $\dot a$ reaches zero after finite time: the closed universe attains a maximum size and cannot expand forever. This is <recollapse of a closed Friedmann universe with nonnegative pressure>.
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